Verify each of the equalities in (2.1.4).

## Section 2.1 Linear Operations

Let \(\R\) denote the set of real numbers. The (Cartesian) plane is the set

\begin{equation*}
\R^2 = \{(a,b)\colon a,b\in \R\}
\end{equation*}

of ordered pairs of real numbers.

The sum \((a,b)+(c,d)\) of two ordered pairs \((a,b)\) and \((c,d)\) is defined by

\begin{equation}
(a,b)+(c,d) = (a+c,b+d).\tag{2.1.1}
\end{equation}

The scalar multiple \(\alpha(a,b)\) of the real number \(\alpha\) times the ordered pair \((a,b)\) is defined by

\begin{equation}
\alpha(a,b) = (\alpha a,\alpha b).\tag{2.1.2}
\end{equation}

The dot product \((a,b)\cdot (c,d)\) of two ordered pairs \((a,b)\) and \((c,d)\) is defined by

\begin{equation}
(a,b)\cdot (c,d) = ac+bd.\tag{2.1.3}
\end{equation}

An ordered pair is sometimes called a

*point*in \(\R^2\text{,}\) and sometimes called a*vector*, depending on the nature of the object represented by \((a,b)\text{.}\) In drawings, a point is a dot, and a vector is an arrow. We use capital letters in italic font to denote points, for example, \(P=(a,b)\text{.}\) We use a bold lower case letters in Roman font, for example \({\mathbf x}=(a,b)\) (better for typing), or lower case letters with an arrow written above them, for example \(\vec{x}=(a,b)\) (better for handwriting), to denote vectors. Given points \(Q=(x_0,y_0)\) and \(R=(x_1,y_1)\text{,}\) we write \(\overrightarrow{QR}\) to denote the ordered pair \(\overrightarrow{QR}= R-Q = (x_1-x_0,y_1-y_0)\text{,}\) which we interpret as a vector, depicted by a directed line segment that begins at \(Q\) and ends at \(R\text{.}\) Note that the*point*\((a,b)\) has only one possible drawing, that is, a dot at the location \((a,b)\text{.}\) By contrast, the*vector*\((a,b)\) has infinitely many possible drawings: given any point \(S\text{,}\) vector \((a,b)\) is depicted by an arrow that begins at \(S\) and ends at \(S+(a,b)\text{.}\) Another way to say this is that two arrows \(\overrightarrow{QR}, \overrightarrow{Q'R'}\) are*equal as vectors*if \(R-Q=R'-Q'\text{,}\) even though we may have \(Q\neq Q'\) and \(R\neq R'\text{.}\)*Warning about notation:*Many texts make a distinction between points and vectors by using the notation \((a,b)\) for a point and \(\langle a,b\rangle\) for a vector. In these notes, we use \((a,b)\) for both the point and the vector. The point \(P=(a,b)\) and the vector \(\mathbf{x}=(a,b)\) are

*not*equal, because a point and a vector are different types of objects. The correct relationship between \(P\) and \(\mathbf{x}\) is given by \(\mathbf{x}=\overrightarrow{OP}\text{,}\) where \(O\) is the point \(O=(0,0)\text{.}\) The take-away message is that you must always be clear whether you mean a point or a vector when you write an ordered pair.

The ordered pair \((0,0)\) is called the origin or the zero vector, depending on whether we are thinking of \((0,0)\) as a point or as a vector, respectively. It is common practice to use the capital letter \(O\) to denote the origin and to use the symbols \(\mathbf{0}\) or \(\vec{0}\) to denote the zero vector.

\begin{align*}
O \amp = (0,0) \amp\amp \text{(the origin point)}\\
\mathbf{0} =\vec{0} \amp = (0,0) \amp\amp \text{(the zero vector)}
\end{align*}

The vectors \({\mathbf e}_1 = (1,0)\) and \({\mathbf e}_2= (0,1)\) are called the standard basis vectors. We have the following expressions for \({\mathbf x}=(a,b)\) in terms of the standard basis vectors.

\begin{equation}
{\mathbf x} = (a,b) = a{\mathbf e}_1 + b{\mathbf e}_2=
({\mathbf e}_1\cdot {\mathbf x}) {\mathbf e}_1
+ ({\mathbf e}_2\cdot {\mathbf x}) {\mathbf e}_2.\tag{2.1.4}
\end{equation}

###
Checkpoint 2.1.1.

The norm or modulus of an ordered pair \((a,b)\text{,}\) denoted \(\left\Vert(a,b)\right\Vert\) , is defined by

\begin{equation}
\left\Vert(a,b)\right\Vert = \sqrt{(a,b)\cdot (a,b)} = \sqrt{a^2+b^2}.\tag{2.1.5}
\end{equation}

The unit circle, denoted \(S^1\) , is the set of all ordered pairs in the Cartesian plane with norm equal to one. Geometrically, \(S^1\) is a circle with radius 1, centered at the origin \((0,0)\text{.}\)

\begin{equation*}
S^1 = \{(x,y)\in \R^2\colon x^2+y^2=1\}
\end{equation*}

It is useful to navigate on the unit circle with a single real parameter, where the parameter value 0 corresponds to the point \((1,0)\text{.}\) A parameter value \(t\gt 0\) corresponds to the point \(W(t)\) that lies at the end of a counterclockwise arc along \(S^1\) that begins at \((1,0)\) and has length \(t\text{.}\) For \(t\lt 0\text{,}\) the point \(W(t)\) lies at the end of a

*clockwise*arc that begins at \((1,0)\) and has length \(|t|\text{.}\) Another way to say this is that \(t\) is the radian measure of the oriented angle \(\angle SOP\text{,}\) where \(S=(1,0)\) and \(O=(0,0)\text{,}\) and \(P=\wrap(t)\text{.}\) The function \(\wrap\colon \R\to S^1\) (often called the wrapping function) is given by
\begin{equation}
\wrap(t)= (\cos t,\sin t)\tag{2.1.6}
\end{equation}

where \(t\) is in radians. Note that \(\wrap\) is periodic: the circumference of the unit circle is \(2\pi\text{,}\) so we have \(\wrap(t)=\wrap(t+2\pi k)\) for every integer \(k\text{.}\)

###
Checkpoint 2.1.2.

Draw a figure that illustrates the points \(P=W(t),S,O\text{,}\) the angle \(\angle SOP\text{,}\) and the oriented arc from \(S\) to \(P\) on \(S^1\text{.}\) Draw one figure for a positive value of \(t\text{,}\) and draw another figure for a negative value of \(t\text{.}\)

Every point \(P\neq (0,0)\) is a positive scalar multiple of a unique point \(\wrap(t)\) on the unit circle, as follows. Given an arbitrary point \(P\in \R^2\) with \(P\neq (0,0)\text{,}\) let \(P'\) be the intersection of the line segment \(\overline{OP}\) with the unit circle, so that we have \(P'=\wrap(t)\) for some real number \(t\text{.}\) It is easy to check that \(P=rP'\) where \(r=\left\Vert P\right\Vert\text{.}\) The expression \(P=r\wrap(t)\) is called the polar form for \(P\text{.}\) The numbers \(r,t\) are called polar coordinates for \(P\text{.}\) (By contrast, the numbers \(a,b\) are called the Cartesian coordinates or the rectangular coordinates for \(P=(a,b)\text{.}\)) The number \(r>0\) is the norm of \(P\text{,}\) and the number \(t\) is called (an) argument of \(P\text{,}\) denoted \(\arg(P)\text{.}\)

\begin{equation}
\arg(r\cos t, r\sin t) = t\tag{2.1.7}
\end{equation}

Because \(\wrap\) is periodic with period \(2\pi\text{,}\) any number of the form \(t+2\pi k\text{,}\) for all \(k\in \Z\text{,}\) is also a value of \(\arg(P)\text{.}\) In this usage, \(\arg\) is not technically a function, because it has multiple possible values. A single-valued function may be defined by restricting the value of \(\arg(P)\text{,}\) say, to the range \(-\pi \lt \arg(P)\leq \pi\text{.}\)

###
Checkpoint 2.1.3.

Show that "it is easy to check" that \(P=rP'\) by checking it.

Rotations about a point in the plane combine in a simple way: a rotation by \(t\) radians, followed by a rotation by \(s\) radians, results in a combined rotation by \(t+s\) radians. It is useful to encode the composition of rotations in a multiplication operation on \(S^1\text{.}\) Given points \(\wrap(t),\wrap(s)\) on the unit circle, let \(\wrap(t)\odot \wrap(s)\) be defined by

\begin{equation}
\wrap(t)\odot \wrap(s) = \wrap(t+s).\tag{2.1.8}
\end{equation}

### Exercises Exercises

#### 1.

Let \(P=(a,b)\) be a point in \(\R^2\) different from the origin \(O=(0,0)\text{,}\) and let \(\alpha \) be a real number. Show that \(O,P,\alpha P\) are collinear (that is, show that \(O,P,\alpha P\) lie on the same line).

#### 2. Parallelogram rule for vector addition.

Let \(P=(a,b)\) and \(Q=(c,d)\) be points in \(\R^2\) such that both \(P\) and \(Q\) are different from the origin \(O=(0,0)\) and \(Q\) is not a scalar multiple of \(P\text{.}\) Let \(R=P+Q\text{.}\) Show that the points \(O,P,Q,R\) are the corners of a parallelogram.

^{ 1 }

A four-sided polygon with consecutive vertices \(A,B,C,D\) is a parallelogram if both pairs of opposite sides are parallel, that is, if \(\overline{AB}\parallel \overline{CD}\) and \(\overline{BC}\parallel \overline{AD}\text{.}\)

*Comment:*This result is called the "parallelogram rule for vector addition" because it gives a simple geometric picture: the sum \(\mathbf{u}+\mathbf{v}=\overrightarrow{OR}\) of vectors \(\mathbf{u}=\overrightarrow{OP},\mathbf{v}=\overrightarrow{OQ}\) is a diagonal of the parallelogram with adjacent sides \(\overline{OP},\overline{OQ}\text{.}\) Make a sketch!

#### 3.

Show that the following hold for all \(\alpha,\beta \in
\R\text{,}\) \({\mathbf x},{\mathbf y}\in \R^2\text{.}\)

\begin{align}
\alpha({\mathbf x}+{\mathbf y}) \amp = \alpha{\mathbf x} +
\alpha{\mathbf y}\tag{2.1.9}\\
(\alpha \mathbf{x})\cdot (\beta \mathbf{y}) \amp = (\alpha \beta) (\mathbf{x}\cdot\mathbf{y})\tag{2.1.10}
\end{align}

#### 4.

How do norms and absolute values interact?

- Show that \(\left\Vert r(a,b)\right\Vert=|r|\left\Vert(a,b)\right\Vert\) for all real numbers \(r,a,b\text{.}\)
- Is it true that \(|(a,b)\cdot (c,d)|= \left\Vert(a,b)\right\Vert\;\left\Vert(c,d)\right\Vert\) for all real numbers \(a,b,c,d\text{?}\) Prove it or give a counterexample.

#### 5. Properties of the wrapping function.

- Show that \(\wrap(t)\cdot \wrap(s) = \cos (t-s)\text{.}\)
- Show that\begin{equation*} \wrap(t)\cdot \wrap(s) = (\wrap(r)\odot \wrap(t))\cdot (\wrap(r)\odot \wrap(s)) \end{equation*}for all real numbers \(r,s,t\text{.}\)

#### 6.

Write formulas for converting from polar to rectangular coordinates and vice-versa.

#### 7. Geometric content of the dot product.

Let \({\mathbf x},{\mathbf y}\) be vectors in \(\R^2\text{.}\) Let \(O=(0,0)\) denote the origin, and let \(P,Q\) be the points given by \({\mathbf x}=\overrightarrow{OP}\) and \({\mathbf y}=\overrightarrow{OQ}\text{.}\) Show that

\begin{equation}
{\mathbf x} \cdot {\mathbf y} = \left\Vert{\mathbf
x}\right\Vert \left\Vert{\mathbf y}\right\Vert \cos \theta\tag{2.1.11}
\end{equation}

where \(\theta\) is the measure of the angle \(\angle POQ\text{.}\) Hint: Start by writing \(P,Q\) in polar form, then use previous exercises.

#### 8.

Let \(n\) be a positive integer. For \(k\) in the range \(0\leq k\leq n-1\text{,}\) let \(P_k=\wrap\left(\frac{2\pi k}{n}\right)\text{,}\) where \(\wrap\) denotes the wrapping function.

- Show that the points \(P_0,P_1,\ldots,P_{n-1}\) are vertices of a regular \(n\)-gon inscribed in the unit circle.
- Show that the following holds for \(0\leq k,\ell\leq n-1\text{.}\)\begin{equation*} P_k\odot P_\ell = P_{k+_n \ell} \end{equation*}

#### 9. Conventions on linear operations with points and vectors.

In these notes, addition and scaling operations are defined for ordered pairs, regardless of whether we view the ordered pairs as points or as vectors. Traditional convention places some restrictions that we have ignored, and will continue to ignore, but nevertheless will present here for the reader’s cultural awareness. In short, traditional orthodoxy goes like this:

- The operations \((a,b)+(c,d)\) and \(\alpha(a,b)\) are allowed when \((a,b)\) and \((c,d) \) are both
*vectors*, and the resulting ordered pairs \((a+c,b+d)\text{,}\) \((\alpha a,\alpha b)\) are interpreted as*vectors*. - Addition and scaling of ordered pairs are discouraged, or even banned outright, if the ordered pairs are both
*points*, except for the following two cases: - If \(P=(a,b)\) and \(Q=(c,d)\) are
*points*, then the operation \((c,d)-(a,b)\) is allowed, and the resulting ordered pair \(Q-P=(c-a,d-b)\) is interpreted as a*vector*, denoted \(\overrightarrow{PQ}\text{.}\) - If \((a,b)\) is a
*point*and \((c,d)\) is a*vector*, then the operation \((a,b)+(c,d)\) is allowed, and the resulting ordered pair is interpreted as a*point*.

Here is a very short summary of the orthodox rules.

- vector \(+\) vector \(=\) vector
- point \(+\) vector \(=\) point
- point \(-\) point \(=\) vector
- scalar \(\cdot \) vector \(=\) vector
- point \(+\) point \(=\) (not recognized)
- scalar \(\cdot \) point \(=\) (not recognized)

The reader will have already noticed that in these notes, we adopt an inclusive approach that allows addition and scaling of points.

- Draw sketches that illustrate the following.
- (vector \(+\) vector \(=\) vector)\begin{equation*} \overrightarrow{PQ}+\overrightarrow{QR}=\overrightarrow{PR} \end{equation*}
- (point \(+\) vector \(=\) point)\begin{equation*} P+\overrightarrow{PQ}=Q \end{equation*}
- (point \(-\) point \(=\) vector)\begin{equation*} Q-P=\overrightarrow{PQ} \end{equation*}

- Let \(P=(a,b)\) be a point, let \(\mathbf{x}\) be a vector, and let \(O=(0,0)\) be the origin. Draw sketches that demonstrate the following "conversion" formulas that convert the point \((a,b)\) to the vector \((a,b)\text{,}\) and vice-versa.
- (convert \(P\) to \(\mathbf{x}\))\begin{equation*} \mathbf{x}=P-O \end{equation*}
- (convert \(\mathbf{x}\) to \(P\))\begin{equation*} P=O+\mathbf{x} \end{equation*}